Problem: Drazil likes heap very much. So he created a problem with heap:

There is a max heap with a height h implemented on the array. The details of this heap are the following:

This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].

Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:

Choose an index i, which contains an element and call the following function f in index i:

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Note that we suppose that if a[i]=0, then index i don't contain an element.

After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.

Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].

The total sum of n is less than 2^{20}.

Output

For each test case, print two lines.

The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.

Example

Input


2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1


Output


10
3 2 3 1
8
2 1 3 1

Solution (in JAVA):

Solution:
/*
     D    A    R    K    L      _                 _         O    R    D    D    A
         K                    / |                 | \        L    O    R         D
        A         R         _ / |                 | \ _       K    L         O    R
  D         A    R    K     _ / |         _       | \ _        L    O    R    D    D
      R    K    L       _ / _   |       _ | \\    |   _ \       O    R    D    D
     R         K        _ / _   |       _ | \ \   |   _ \        L    O    R
    D    A    R     _ / _ / |   /     /         \ \   | \ _ \ _        K    L    O
   D    D           _ / _ / |   /     /          \ \   | \ _ \ _             A
       K        _ / _ /     | /     /     _       \ \ |     \ _ \ _      L    O    R
 D    A        _ / _ /      | /     /     _       \ \ |      \ _ \ _      R         K
            _ / _ /         | |     |   |   \ (O) / | |         \ _ \ _         O
    D      _ / _ /          | |     |   |    \   /  | |          \ _ \ _         D
   A      / _ /     |   |   | \     |   \ _   \|/   / |   |   |     \ _ \
  K      / _ /      |   |   | \     |   \ _    V    / |   |   |      \ _ \         L
        / /         |   |   |   \ _ /       \ _ _ /   |   |   |         \ \
       / /          |   |   |                         |   | | |          \ \
      / /       |   | | \     \       ROWECHEN      /     / | |   |       \ \
     / /    _ _ |   | | \     \        ZHONG        /     / | |   | _ _    \ \
    / / _ /   \ | | | |   \ / \ \                 / / \ /   | | | | /   \ _ \ \
   / / _ /    \ | | | |       \ \                 / /       | | | | /    \ _ \ \
  / / /         \ \ \ \ /       /                 \       \ / / / /         \ \ \
 \ / /          \ \ \ \ /       /                 \       \ / / / /          \ \ /
  \|/             \|/         |         |         |           \|/             \|/
  L         O     \|/         |         |   |     |           \|/        R    D
 A    R    K    L             / _ |     |   | _     \      O    R    D    D    A    R
     L    O         R         / _ |     |   | _     \       D    D         A    R
    L    O    R    D        / / / _     | _ | | \ _   \      D    A    R         K
   O    R    D              / / / _ |   | _ | | \ _   \       D    A    R    K    L
  R    D    D    A    R     | / / | | / | | \ / | | \ |        K    L    O    R    D
 A         R         K      | / / | | / | | \ / | | \ |              L    O         R
     D    A         R         /   \ / | | | / | | / \ /     K    L         O    R
         D    D         A     /   \ / | | | / | | / \/            R    K    L    O
             R         D    D     / | / \ |   \ /        A    R    K    L    O    R
       D    A    R    K    L      / | / \ |    \/              O    R    D    D
 R    K    L    O    R    D       \ /     |      D    A    R         K    L    O    R
          D    A    R             \/      |            K    L    O    R    D    D

*/
//TEMPLATE V2
import java.io.*;
import java.util.*;

public class Main {
//Solution goes below: ------------------------------------


static int[] a;
static boolean[] fixed;
static int g;
static int cap;
public static boolean remove(int i){
    int l = 2*i;
    int r = 2*i+1;
    boolean res;
    if(a[l] ==0&&a[r]==0) if (res = i >= cap) a[i] = 0;
    else fixed[i] = true;
    else if (a[l] > a[r]) {
        int t = a[l];
        if (res = remove(l)) a[i] = t;
        else fixed[i] = true;
    } else {
        int t = a[r];
        if (res = remove(r)) a[i] = t;
        else fixed[i] = true;
    }
    return res;
}

public static void solution() throws IOException{
    int t = nextInt();
    for(int test=0;test<t;test++){
        int h = nextInt();
        g= nextInt();
        int size = 1<<h;
        cap = 1<<g;
        int req = size-cap;

        a = new int[2*size+2];//Two times so it doesn't out-of-bounds.
        for(int i=1;i<size;i++) a[i] = nextInt();

        //fixed is the elements that are finalized and cannot move anymore.
        fixed = new boolean[size];
        int curr = 1;

        int[] res = new int[req];
        int j=0;
        while(j<req){
            while(curr<size&&fixed[curr]) curr++;
            if(remove(curr)) res[j++] = curr;
        }
        long total=0;
        for(int i=1;i<cap; i++) total += a[i];
        println(total);
        for(int i:res){
            print(i);print(" ");
        }
        println();
    }


}

//Solution goes above: ------------------------------------
//IO
public static BufferedReader br;
public static StringTokenizer st;
public static BufferedWriter bw;
public static void main(String[] args) throws IOException{
    br = new BufferedReader(new InputStreamReader(System.in));
    bw = new BufferedWriter(new OutputStreamWriter(System.out));
    solution();
    bw.close();//Flushes too.
}

public static String nextToken() {
    while (st == null || !st.hasMoreElements()) try {
        st = new StringTokenizer(br.readLine());
    } catch (IOException e) {
        e.printStackTrace();
    }
    return st.nextToken();
}
public static int nextInt() {
    return Integer.parseInt(nextToken());
}

public static void println(Object s) throws IOException{
    bw.write(s.toString()+"\n");
}

public static void println() throws IOException{
    bw.newLine();
}

public static void print(Object s) throws IOException{
    bw.write(s.toString());
}
}